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KINEMATICS ANALYSIS
OF ROBOTS(Part 4)
This lecture continues the discussion on the analysis of the forward and inverse kinematics of robots.
After this lecture, the student should be able to:•Solve problems of robot kinematics analysis using transformation matrices
Kinematics Analysis of Robots IV
Example: A 3 DOF RRR Robot
Inverse Kinematics if no restriction on the orientation of point “P”. This is useful if we are concerned only on the position of “P”
Y0, Y1
X0, X1
Z0, Z1
Z2
X2
Y2
Z3
X3
Y3
A=3 B=2 C=1
P
Example: A 3 DOF RRR Robot
In this case we are not given the following matrix for the description of the orientation and position of point “P”:
TTTTpaon
paon
paon
zzzz
yyyy
xxxx
03
23
12
01
1000
Instead, we need to find 1, 2, and 3 given only the location of “P” w.r.t. to the base frame:
1
1
0
z
y
x
p
p
p
p
Example: A 3 DOF RRR Robot
11
303
0 pT
p
We know that
1
)cos()sin(
)]cos()cos()[sin(
)]cos()cos()[cos(
1
1
0
0
1000
)sin(0)cos()sin(
)sin())cos(()cos()sin()sin()cos()sin(
)cos())cos(()sin()sin()cos()cos()cos(
1
322
3221
3221
3
0
3
23232
121321321
121321321
3
0
3
z
y
x
p
p
p
CB
CBA
CBA
pT
C
B
BA
BA
pT
Example: A 3 DOF RRR Robot
Equate elements:
)cos()cos()cos(
)]cos()cos()[cos(
3221
3221
CBA
p
CBAp
x
x
)cos()cos()sin(
)]cos()cos()[sin(
3221
3221
CBA
p
CBAp
y
y
x
y
p
p11 tan
Be Careful!
Example: A 3 DOF RRR Robot
Be careful when using your calculator to find inverse tangent!
y
xxy)tan(
1st quadrant (x=+ve, y=+ve)
2nd quadrant (x=-ve, y=+ve)
4th quadrant (x=+ve, y=-ve)
3rd quadrant (x=-ve, y=-ve)
Your calculator can only give the angle in the 1st quadrant. You have to adjust the answer from the calculator
Example: A 3 DOF RRR Robot
)cos()cos()cos(
)]cos()cos()[cos(
3221
3221
CBAp
CBAp
x
x
)cos()cos()sin(
)]cos()cos()[sin(
3221
3221
CBAp
CBAp
y
y
If cos(1)0, let Apx
)cos( 1
Otherwise, let Apy
)sin( 1
Example: A 3 DOF RRR Robot
We now have2
3223222
222
322
)cos()cos(2)(cos)(cos
)cos()cos(
BCCB
CB
From the comparison of the 3rd row elements:
232232
222
22
322
)sin()sin(2)(sin)(sin
)sin()sin(
z
z
pBCCB
pCB
Combining these equations, we get
BCCBp
pBCCB
z
z
2)cos(
)cos(22222
3
223
22
)cos(
)sin(tan
)(cos1)sin(
3
313
32
3
Example: A 3 DOF RRR Robot
)sin()sin()cos()cos()cos(
)sin()cos()cos()sin()sin(
212121
212121
)sin()]sin([)cos()]cos([
)]sin()sin()cos()[cos()cos(
)cos()cos(
2323
32322
322
CCB
CB
CB
Using
z
z
z
pCCB
pCB
pCB
)cos()]sin([)sin()]cos([
)]sin()cos()cos()[sin()sin(
)sin()sin(
2323
32322
322
Let )]sin([)],cos([ 33 CbCBa
zpba
ba
)cos()sin(
)sin()cos(
22
22
Example: A 3 DOF RRR Robot
zpba
ba
)cos()sin(
)sin()cos(
22
22
We can now solve the above equations to get
222 )sin(babapz
If a0, thenab )sin(
)cos( 22
If b0, thenbapz )sin(
)cos( 22
)cos()sin(
tan2
212
Example: A 3 DOF RRR Robot
Y0, Y1
X0, X1
Z0, Z1
Z2
X2
Y2
Z3
X3
Y3
A=3 B=2 C=1
P
1
1
0
5
1
0 p
Now find 1, 2, and 3 to move point “P” to
Example: A 3 DOF RRR Robot
0tan5
0 11
x
y
x
y
p
p
p
p
Now cos(1)0, let 2)cos( 1
A
px
02
)cos(2222
3 BC
CBpz
90)cos(
)sin(tan
1)(cos1)sin(
3
313
32
3
I.e. Assume elbow down
Example: A 3 DOF RRR Robot
1)]sin([
2)]cos([
3
3
Cb
CBa
05
22)sin( 222
babapz
Now a0, then 1)sin(
)cos( 22
ab
0)cos()sin(
tan2
212
Compare this with the results form the previous lecture! The difference is that now we are not concerned with the orientation of point “P”.
Example: A 3 DOF RPR Robot
Link and Joint Assignment
Link (0)
Link (1)Link (2) Link (3)
Revolute joint <1>
Prismatic joint <2>
Revolute joint <3>
Example: A 3 DOF RPR Robot
Frame Assignment
Z1
Z1
X1
X1
Y1
Example: A 3 DOF RPR Robot
Frame Assignment
Z0, Z1
Y0, Y1
X0, X1
Z2Z2
X2
Y2
Y2
Example: A 3 DOF RPR Robot
Frame Assignment
Z0, Z1
Y0, Y1
X0, X1
Z2
Y2
X2
Z3
X3
Y3
Y3
Example: A 3 DOF RPR Robot
Frame Assignment
Z0, Z1
Y0, Y1
X0, X1
Z2
Y2
X2
Z3
Y3
X3
1
d2 3
Example: A 3 DOF RPR Robot
Tabulation of D-H parametersZ0, Z1
Y0, Y1
X0, X1
Z2
Y2
X2
Z3
Y3
X3
0 = (angle from Z0 to Z1 measured along X0) = 0°a0 = (distance from Z0 to Z1 measured along X0) = 0d1 = (distance from X0 to X1 measured along Z1)= 01 = variable (angle from X0 to X1 measured along Z1)1 = 0° (at home position) but 1 can change as the arm moves
Example: A 3 DOF RPR Robot
Tabulation of D-H parametersZ0, Z1
Y0, Y1
X0, X1
Z2
Y2
X2
Z3
Y3
X3
1 = (angle from Z1 to Z2 measured along X1) = -90°a1 = (distance from Z1 to Z2 measured along X1) = 0d2 = variable (distance from X1 to X2 measured along Z2)2 = variable (angle from X1 to X2 measured along Z2) = 180°
Example: A 3 DOF RPR Robot
Tabulation of D-H parametersZ0, Z1
Y0, Y1
X0, X1
Z2
Y2
X2
Z3
Y3
X3
2 = (angle from Z2 to Z3 measured along X2) = 0°a2 = (distance from Z2 to Z3measured along X2) = 0d3 = (distance from X2 to X3 measured along Z3) = A3 = variable (angle from X2 to X3 measured along Z3)3= 0° (at home position) but 3 can change as the arm moves
A
Link i Twist i Link length ai
Link offset di
Joint angle i
i=0 0 0 … …
i=1 -90° 0 0 1
(1=0° at home position)
i=2 0 0 d2 2=180°
i=3 … … A 3
(3=-0° at home position)
Summary of D-H parameters
Example: A 3 DOF RPR Robot
Tabulation of Transformation Matrices from the D-H table
1000
)cos()cos()cos()sin()sin()sin(
)sin()sin()cos()cos()sin()cos(
0)sin()cos(
1111
1111
1
1
iiiiiii
iiiiiii
iii
ii d
d
a
T
0,0,0 100 da
1000
0100
00)cos()sin(
00)sin()cos(
11
11
01
T
Example: A 3 DOF RPR Robot
Tabulation of Transformation Matrices from the D-H table 180,0,90 211 a
1000
0010
100
0001
212
dT
Ada 322 ,0,0
1000
100
00)cos()sin(
00)sin()cos(
33
33
23 AT
Example: A 3 DOF RPR Robot
Once all the transformation matrices are obtained, you can then proceed to get the overall transformation matrix for the forward kinematics.
After that, given the position and orientation of the point “P” on the gripper, you can proceed to compare the terms of the matrices to get the inverse kinematics. For the inverse kinematics, you will be solving for 1, d2, and 3
Try it out as Homework.
Summary
This lecture continues the discussion on the analysis of the forward and inverse kinematics of robots.
The following were covered:•Problems of robot kinematics analysis using transformation matrices
